Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(a(f(b, a(x)))) → A(a(x))
A(a(x)) → F(a(x), b)
A(a(x)) → A(f(a(x), b))
A(a(x)) → F(b, a(f(a(x), b)))
A(a(f(b, a(x)))) → A(a(a(x)))
A(a(f(b, a(x)))) → F(a(a(a(x))), b)
F(a(x), b) → F(b, a(x))
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(b, a(x)))) → A(a(x))
A(a(x)) → F(a(x), b)
A(a(x)) → A(f(a(x), b))
A(a(x)) → F(b, a(f(a(x), b)))
A(a(f(b, a(x)))) → A(a(a(x)))
A(a(f(b, a(x)))) → F(a(a(a(x))), b)
F(a(x), b) → F(b, a(x))
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(a(f(b, a(x)))) → A(a(x))
A(a(x)) → A(f(a(x), b))
A(a(x)) → F(a(x), b)
A(a(x)) → F(b, a(f(a(x), b)))
A(a(f(b, a(x)))) → A(a(a(x)))
A(a(f(b, a(x)))) → F(a(a(a(x))), b)
F(a(x), b) → F(b, a(x))
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
A(a(f(b, a(x)))) → A(a(x))
A(a(f(b, a(x)))) → A(a(a(x)))
The TRS R consists of the following rules:
a(a(f(b, a(x)))) → f(a(a(a(x))), b)
a(a(x)) → f(b, a(f(a(x), b)))
f(a(x), b) → f(b, a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.